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Unformatted text preview: University of California, Berkeley, Statistics 134: Concepts of Probability Michael Lugo, Spring 2011 Exam 1 February 16, 2011, 11:10 am  12:00 noon Name: Solutions Student ID: This exam consists of seven pages: this cover page; five pages, each containing one problem with several parts; and a table of the normal distribution. You may use a calculator, and notes on one sides of a standard 8.5by11inch sheet of paper which you have written by hand, yourself. You must show all work other than basic arithmetic. Write your name at the top of each page. Each problem is worth 20 points, for a total of 100. Weights of problem subparts are indicated on the problems. DO NOT WRITE BELOW THIS LINE 1. 2. 3. 4. 5. TOTAL 1 Name: Solutions 1. Three ordinary, sixsided dice are rolled. (Each has the numbers 1, 2, 3, 4, 5 and 6 on its faces.) What is the probability that: (a) All three dice show different numbers? [5 points] 6 / 6 5 / 6 4 / 6 = 120 / 216 = 5 / 9. The probability that the second die doesnt show the same number as the first die is 5 / 6; given that this happens, the probability that the third die doesnt show the same number as either of the second two dice is 4 / 6. (b) The sum of the numbers showing on the dice is 16? [5 points] 16 can be obtained as two 5s and one 6, or one 4 and two 6s. Either of these can occur in three ways: (5 , 5 , 6) , (5 , 6 , 5) , (6 , 5 , 5) , (4 , 6 , 6) , (6 , 4 , 6) , (6 , 6 , 4). There are a total of 6 3 = 216 possible outcomes for rolling three dice, so the answer is 6 / 216 or 1 / 36. (c) The largest of the three numbers shown is 5? [5 points] Let L n be the event that the largest number shown is n , and let M n be the event that the largest number shown is less than or equal to n . Then P ( M n ) = ( n/ 6) 3 , since for the largest number shown to be at most n , all three numbers shown must be at most n . Now, L 5 = M 5 M c 4 for the largest number shown to be 5, it must be less than or equal to 5, but not less than or equal to 4. So P ( L 5 ) = P ( M 5 M c 4 ) = P ( M 5 ) P ( M 5 M 4 ). But M 5 M 4 is just M 4 . So the answer is P ( M 5 ) P ( M 4 ) = (5 3 4 3 ) / 6 3 = 61 / 216....
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 Fall '11
 Shobhana
 Statistics, Normal Distribution, Probability, Dice, Probability space

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